Example 3.4.1 – Application to Sequences | 例3.4.1

📚 Example 3.4.1 – Application to Sequences | 例3.4.1 – 数列的应用

Sequences are one of the most versatile topics in A-Level Mathematics. They appear not only as abstract mathematical objects but also as powerful tools for modelling real‑world situations such as financial growth, depreciation, population dynamics, and even the arrangement of objects in a pattern. In this article, we will study a classic exam‑style problem — Example 3.4.1 — which demonstrates how arithmetic and geometric sequences can be applied to a practical context. We will break down the solution step by step, highlight key methods, and provide revision tips that will help you tackle similar questions with confidence.

数列是A-Level数学中最灵活的主题之一。它们不仅是抽象的数学对象，更是建模现实世界情况的有力工具，例如金融增长、折旧、人口动态，甚至物体排列的规律。在本文中，我们将研究一道经典的考试型问题——例3.4.1——它展示了等差数列和等比数列如何应用于实际情境。我们将逐步分解解答过程，强调关键方法，并提供复习技巧，帮助你自信应对类似问题。

1. Introduction to Sequences | 数列简介

A sequence is an ordered list of numbers, usually written as a₁, a₂, a₃, … . Each number in the sequence is called a term. The subscript indicates the position of the term — for instance, aₙ is the nth term. Sequences can be defined by a formula that allows us to calculate any term directly, or by a recurrence relation that connects each term to the previous one(s). In the AQA specification, you are expected to understand both arithmetic and geometric sequences, as well as their sums (series).

数列是按一定顺序排列的数表，通常写作a₁, a₂, a₃, …。数列中的每一个数被称为“项”。下标表示该项的位置——例如，aₙ 是第 n 项。数列可以通过公式直接计算任意一项，也可以通过递推关系式将每一项与其前一项（或前几项）联系起来。在AQA考试大纲中，要求掌握等差数列和等比数列及其求和（级数）。

The application of sequences often involves translating a worded problem into a mathematical model. This requires identifying the type of sequence, extracting key values, and then performing the necessary calculations. The example we will explore in this article is a perfect illustration of this process.

数列的应用通常涉及将文字问题转化为数学模型。这需要识别数列的类型、提取关键数值，然后进行必要的计算。我们将在本文中探讨的例题正是这种过程的完美例证。

2. Arithmetic Sequences Recap | 等差数列复习

An arithmetic sequence is one in which the difference between consecutive terms is constant. This constant is called the common difference, denoted by d. The nth term of an arithmetic sequence is given by:

等差数列是指相邻两项之差为常数的数列。这个常数称为公差，用 d 表示。等差数列的第 n 项公式为：

aₙ = a₁ + (n − 1)d

where a₁ is the first term. The sum of the first n terms, Sₙ, can be found using either of the following equivalent formulas:

其中 a₁ 是首项。前 n 项的和 Sₙ 可以用下面两个等价公式之一计算：

Sₙ = n/2 [2a₁ + (n − 1)d] = n/2 (a₁ + aₙ)

Arithmetic sequences are used to model situations where a quantity changes by a fixed amount each step. For example, a taxi fare that increases by a fixed amount per extra kilometre, or the number of seats in successive rows of a theatre.

等差数列用于建模每步变化量为固定值的情形。例如，每增加一公里固定加价的出租车费用，或剧院中连续排座位的数量。

3. Geometric Sequences Recap | 等比数列复习

A geometric sequence is one in which each term is obtained by multiplying the previous term by a fixed, non‑zero constant called the common ratio, denoted by r. The nth term is given by:

等比数列是指每一项都通过乘一个固定且非零的常数（称为公比，用 r 表示）得到前一项的数列。第 n 项公式为：

aₙ = a₁rⁿ⁻¹

The sum of the first n terms (for r ≠ 1) is:

前 n 项和（r ≠ 1 时）为：

Sₙ = a₁(rⁿ − 1) / (r − 1)

Geometric sequences are used when a quantity changes by a fixed percentage or ratio. Examples include compound interest, population growth, radioactive decay, and the depreciation of an asset over time.

等比数列用于建模以固定百分比或比率变化的情形。例如复利、人口增长、放射性衰变，以及资产随时间的折旧。

4. The Context of Example 3.4.1 | 例3.4.1的背景

Example 3.4.1 in the AQA course is designed to show how sequences can be applied to a real‑life financial problem. A common scenario involves the depreciation of a car. Suppose a car is bought for £25,000 and its value falls by 12% every year. We are asked to find the value after a certain number of years, and to determine when the car’s value will first drop below a given amount.

AQA课程中的例3.4.1旨在展示数列如何应用于现实中的财务问题。一个常见的情景是汽车贬值。假设一辆汽车以25,000英镑的价格购入，其价值每年下降12%。题目要求我们计算若干年后的价值，并确定汽车价值首次降至指定金额以下的时间。

This problem is naturally modelled by a geometric sequence because the value after each year is a fixed percentage of the previous year’s value. The common ratio is r = 1 − 0.12 = 0.88. By applying the geometric nth‑term formula, we can find the value at any year, and by using logarithms we can solve for the year when the value reaches a threshold.

这个问题天然适合用等比数列建模，因为每一年的价值都是前一年的固定百分比。公比为 r = 1 − 0.12 = 0.88。通过应用等比数列的第 n 项公式，我们可以求出任一年的价值；通过使用对数，我们可以解出价值达到某一阈值时的年份。

However, a well‑designed exam question often adds a twist. For instance, after a certain year, the rate of depreciation might change, or an additional fixed payment may be involved. This forces us to combine both geometric and arithmetic ideas in one solution. In Example 3.4.1, we will see such a combination.

然而，设计良好的一道考试题往往会增加变化。例如，在某一特定年份之后，贬值率可能改变，或者可能涉及额外的固定付款。这迫使我们在一个解答中同时结合等比和等差的思想。在例3.4.1中，我们将看到这样的结合。

5. Example 3.4.1: Problem Statement | 例3.4.1：问题陈述

Example 3.4.1 — A company buys a machine for £40,000. The value of the machine depreciates by 10% each year for the first 5 years. After 5 years, the machine is subject to a fixed annual maintenance fee of £500, which is added to its recorded “net value” (i.e., value minus accumulated maintenance costs). The net value after n years, for n ≥ 5, is defined as: net value = machine value after n years − 500 × (n − 5). Find:

例3.4.1 — 一家公司以40,000英镑的价格购买了一台机器。机器价值在前5年每年贬值10%。5年后，机器每年产生固定的维护费500英镑，该费用被计入其“净价值”（即价值减去累计维护成本）。对于 n ≥ 5，第 n 年的净价值定义为：净价值 = 第 n 年机器价值 − 500 × (n − 5)。求：

(a) the machine value after 5 years,
(b) the net value after 8 years,
(c) the first year (n > 5) in which the net value falls below £15,000.

(a) 5年后机器的价值；
(b) 8年后的净价值；
(c) 净价值首次低于15,000英镑的年份（n > 5）。

6. Step‑by‑Step Solution: Part (a) | 分步解答：第(a)部分

For the first 5 years, the machine value follows a geometric sequence with a₁ = 40,000 and r = 0.90 (since 10% depreciation means the value retains 90%). The value after 5 years is the 5th term of this sequence.

前5年，机器价值遵循等比数列，其中 a₁ = 40,000，r = 0.90（因为贬值10%意味着保留90%的价值）。5年后的价值是该数列的第5项。

a₅ = 40,000 × (0.90)⁵

Calculate the power: 0.90⁵ = 0.59049. Therefore:

计算幂：0.90⁵ = 0.59049。因此：

a₅ = 40,000 × 0.59049 = 23,619.60

So the machine value after 5 years is £23,619.60. Note that we round money to 2 decimal places only at the final step, to avoid rounding errors in later parts.

因此5年后机器的价值为23,619.60英镑。注意，我们仅在最后一步将金额四舍五入到小数点后两位，以避免在后续部分产生舍入误差。

7. Step‑by‑Step Solution: Part (b) | 分步解答：第(b)部分

After 5 years, the machine continues to depreciate by 10% per year, but now we must also subtract the accumulated maintenance fees. For n = 8, we first need the machine value after 8 years. Since the depreciation pattern is the same geometric sequence, we can find the 8th term.

5年后，机器继续每年贬值10%，但现在我们还必须减去累计维护费。对于 n = 8，我们首先需要求出8年后的机器价值。由于贬值模式是相同的等比数列，我们可以求出第8项。

a₈ = 40,000 × (0.90)⁸

Computing the power: 0.90⁸ = (0.90⁵) × (0.90³) = 0.59049 × 0.729 = 0.43046721 (exactly). Then:

计算幂：0.90⁸ = (0.90⁵) × (0.90³) = 0.59049 × 0.729 = 0.43046721（精确值）。于是：

a₈ = 40,000 × 0.43046721 = 17,218.6884

Now add the maintenance element. For n = 8, the number of years after year 5 is n − 5 = 3. The accumulated maintenance cost is £500 × 3 = £1,500. The net value is:

现在加入维护费元素。对于 n = 8，5年后的年数为 n − 5 = 3。累计维护成本为500英镑 × 3 = 1,500英镑。净价值为：

Net value = 17,218.6884 − 1,500 = 15,718.6884

Rounded to the nearest penny, the net value after 8 years is £15,718.69.

四舍五入到便士，8年后的净价值为15,718.69英镑。

8. Step‑by‑Step Solution: Part (c) | 分步解答：第(c)部分

We need the first year n > 5 such that the net value falls below £15,000. The net value after n years is:

我们需要找到使净价值低于15,000英镑的第一个年份 n > 5。第 n 年的净价值为：

Net value = 40,000(0.90)ⁿ − 500(n − 5)

We set this less than 15,000 and solve for n. Since n is an integer, we can test successive values, but a more elegant method uses a hybrid approach. First, check n = 9 to see the trend.

我们令该表达式小于15,000并解出 n。由于 n 是整数，我们可以逐个测试值，但更优雅的方法是使用混合方式。首先检查 n = 9 以观察趋势。

For n = 9: machine value = 40,000 × (0.90)⁹ = 40,000 × 0.387420489 = 15,496.81956. Accumulated maintenance = £500 × 4 = £2,000. Net value = 15,496.81956 − 2,000 = 13,496.81956. This is already below 15,000. But is n = 8 the first year? From part (b), net value at n = 8 was 15,718.69, which is above 15,000. Therefore the first year is n = 9.

对于 n = 9：机器价值 = 40,000 × (0.90)⁹ = 40,000 × 0.387420489 = 15,496.81956。累计维护费 = 500英镑 × 4 = 2,000英镑。净价值 = 15,496.81956 − 2,000 = 13,496.81956。这已经低于15,000。但 n = 8 是第一个吗？从第(b)部分可知，n = 8 时的净价值为15,718.69，高于15,000。因此第一个年份是 n = 9。

To be rigorous, we can solve analytically. We require 40,000(0.90)ⁿ − 500(n − 5) < 15,000. Rearranging gives 40,000(0.90)ⁿ < 500(n + 25). Divide by 500: 80(0.90)ⁿ < n + 25. This is a transcendental equation; we can solve by inspection for integer n. Testing n = 8 gives LHS = 80 × 0.43046721 = 34.4374, RHS = 33. Inequality: 34.4374 < 33 is false. For n = 9, LHS = 80 × 0.387420489 = 30.9936, RHS = 34. Inequality true. Thus n = 9 is indeed the first year.

为了严谨，我们可以解析求解。要求 40,000(0.90)ⁿ − 500(n − 5) < 15,000。整理得 40,000(0.90)ⁿ < 500(n + 25)。除以500：80(0.90)ⁿ < n + 25。这是超越方程；我们可以通过对整数 n 进行检验来求解。测试 n = 8 时，左边 = 80 × 0.43046721 = 34.4374，右边 = 33。不等式：34.4374 < 33 为假。对于 n = 9，左边 = 80 × 0.387420489 = 30.9936，右边 = 34。不等式成立。因此 n = 9 确实是第一个年份。

9. Interpreting the Results | 结果解读

The solution shows that the geometric model for depreciation must be adjusted for maintenance costs. In part (a) we isolated the pure geometric decay; in part (b) we combined it with a linear (arithmetic) penalty; in part (c) we used a trial‑and‑improvement method that is acceptable for integer‑valued years. Notice that the net value decreases more rapidly after year 5 because the maintenance subtraction is a fixed amount each year, whereas the depreciation alone is a decreasing amount (10% of a smaller value each year).

解答显示了等比模型在折旧中必须考虑维护成本。第(a)部分我们提取了纯等比衰减；第(b)部分我们将其与线性（等差）惩罚结合；第(c)部分我们使用了试值改进法，该方法对整数年份是可行的。注意，5年后净价值下降得更快，因为维护费减去的是固定金额，而单独折旧是逐年减少的金额（每年较小价值的10%）。

This example also demonstrates a key exam skill: breaking down a multi‑part problem into smaller, manageable steps. The first step is always to identify the type of sequence and write down the relevant formula. Then substitute the given values carefully, and finally check whether the answer makes sense in the context.

这个例题还展示了一项关键的考试技能：将多部分问题拆分为更小、可管理的步骤。第一步始终是识别数列类型并写出相关公式。然后仔细代入给定值，最后检查答案在上下文中是否合理。

10. Exam Tips and Common Pitfalls | 考试技巧与常见误区

When working with sequences, students often make avoidable errors. Here are some common pitfalls and how to avoid them.

在处理数列时，学生常犯可避免的错误。以下是一些常见误区及如何避免它们。

Confusing arithmetic and geometric sequences. Always check whether the change is by a fixed amount (geometric) or a fixed ratio (arithmetic). In the example, the 10% depreciation means multiply by 0.9 each year — that is geometric.
Misreading the term number. “After 5 years” corresponds to n = 5, not n = 4. Make sure you index correctly if the initial term is at time zero.
Rounding prematurely. Round only at the final answer. In part (b) we kept the exact decimal 17,218.6884 to avoid an error of a few pounds in part (c).
Forgetting to subtract maintenance for all years after year 5. The expression £500 × (n − 5) requires counting years correctly — here n − 5, not n.
Solving inequalities without checking integer bounds. In part (c) we verified both n = 8 and n = 9 because the inequality could flip due to the linear term.

在数列题中，学生常犯的错误如下。始终检查变化是固定量（等差）还是固定比率（等比）。在例题中，10%的贬值意味着每年乘以0.9——那是等比。“5年后”对应于 n = 5，而不是 n = 4。确保在初始项位于时间零点时正确编制索引。仅最终答案四舍五入。在第(b)部分我们保留了精确小数17,218.6884，以避免在第(c)部分出现几英镑的误差。忘记减去5年后的所有维护费。表达式 £500 × (n − 5) 需要正确计数年份——这里是 n − 5，而不是 n。在解不等式时要检查整数边界。在第(c)部分我们同时验证了 n = 8 和 n = 9，因为线性项可能导致不等式翻转。

11. Practice Questions | 练习问题

To reinforce your understanding, try these similar problems on your own.

为了巩固理解，请自行尝试以下类似问题。

A population of bacteria starts at 2,000 and increases by 8% every hour. (a) Write down the size of the population after 4 hours. (b) How many hours will it take for the population to exceed 10,000? (Give the smallest integer hour.)
An athlete runs 5 km on the first day of training. Each subsequent day she runs 2 km more than the previous day. (a) Find her distance on the 20th day. (b) What is the total distance run over the first 20 days?
A student invests £3,000 in a savings account that pays compound interest of 4% per annum. After 5 years, she withdraws £1,000 and leaves the rest in the account for another 3 years. Find the final amount, correct to the nearest pound.

一个细菌种群初始为2,000，每小时增长8%。(a) 写出4小时后种群的大小。(b) 多少小时后种群将超过10,000？（给出最小整数小时。）

一名运动员第一天训练跑5公里。此后每天比前一天多跑2公里。(a) 求她第20天的距离。(b) 前20天总共跑的距离是多少？

一名学生在年利率4%的复利储蓄账户中存入3,000英镑。5年后，她取出1,000英镑，其余部分再存3年。求最终金额，精确到最接近的英镑。

12. Key Formulas Summary | 关键公式总结

Below is a quick reference of the formulas used in this article, which are also essential for the AQA exam.

下面是本文中用到的公式快速参考，这些公式在AQA考试中也是必不可少的。

Type \| 类型	nth term \| 第n项	Sum of first n terms \| 前n项和
Arithmetic \| 等差	aₙ = a₁ + (n − 1)d	Sₙ = n/2 [2a₁ + (n − 1)d]
Geometric \| 等比	aₙ = a₁rⁿ⁻¹	Sₙ = a₁(rⁿ − 1)/(r − 1) (r ≠ 1)

Also remember: the common ratio r = 1 ± percentage change, where a decrease uses a minus sign. For an increase of 8%, r = 1.08; for a decrease of 12%, r = 0.88.

还要记住：公比 r = 1 ± 百分比变化，其中减少用减号。增长8%时 r = 1.08；减少12%时 r = 0.88。

13. Conclusion | 结论

Example 3.4.1 illustrates the power of sequences in solving real‑world problems. By recognising the underlying geometric and arithmetic structures, we were able to compute values and make predictions with ease. After the first 5 years, the introduction of a fixed maintenance fee made the net value a hybrid of a geometric decay and an arithmetic subtraction. This combination is a favourite of examiners because it tests multiple concepts at once.

例3.4.1展示了数列在解决现实世界问题中的力量。通过识别潜在的等比和等差结构，我们能够轻松计算数值并做出预测。在最初的5年后，引入固定维护费使得净价值成为等比衰减与等差减法的混合。这种组合是考官的最爱，因为它同时测试了多个概念。

To excel in questions on sequences, practise identifying the type of sequence from a word problem, write down the relevant formulas before substituting, and always check your answers for reasonableness. With enough practice, you will find that these problems become routine.

为了在数列题中取得好成绩，练习从文字问题中识别数列类型，在代入前写出相关公式，并始终检查答案的合理性。多加练习，你会发现这些问题变得得心应手。

Keep this article handy as a revision guide, and revisit the worked example step by step before your exam. Good luck!

将本文作为复习指南，在考试前逐步回顾这个例题。祝你好运！

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Type \| 类型	nth term \| 第n项	Sum of first n terms \| 前n项和
Arithmetic \| 等差	aₙ = a₁ + (n − 1)d	Sₙ = n/2 [2a₁ + (n − 1)d]
Geometric \| 等比	aₙ = a₁rⁿ⁻¹	Sₙ = a₁(rⁿ − 1)/(r − 1) (r ≠ 1)

Example 3.4.1 – Application to Sequences | 例3.4.1 – 数列的应用